Question 8 of 69advanced🔧 ApplyNumerical3 marks

Reaction between N2N_2 and O2O_2 takes place as follows:

2N2(g)+O2(g)⇌2N2O(g)2N_2 (g) + O_2 (g) \rightleftharpoons 2N_2O (g)

If a mixture of 0.482 mol N2N_2 and 0.933 mol of O2O_2 is placed in a 10 L reaction vessel and allowed to form N2ON_2O at a temperature for which Kc=2.0×10−37K_c = 2.0 \times 10^{-37}, determine the composition of equilibrium mixture.

Correct Answer

The equilibrium mixture composition is: [N2]=0.0482[N_2] = 0.0482 M, [O2]=0.0933[O_2] = 0.0933 M, and [N2O]=6.58×10−21[N_2O] = 6.58 \times 10^{-21} M.

Exercise: EXERCISES | Q: 6.8 | (Page 42)
For More Understanding

Explanation

The equilibrium constant Kc=2.0×10−37K_c = 2.0 \times 10^{-37} is extremely small, indicating that the reaction proceeds to a negligible extent. Therefore, the concentrations of the reactants N2N_2 and O2O_2 remain virtually unchanged, and the amount of product N2ON_2O formed is practically zero.

Solution Steps

  1. Step 1: Calculate initial concentrations Volume of vessel (VV) = 10 L

    [N2]0=0.482 mol10 L=0.0482 M[N_2]_0 = \frac{0.482 \text{ mol}}{10 \text{ L}} = 0.0482 \text{ M} [O2]0=0.933 mol10 L=0.0933 M[O_2]_0 = \frac{0.933 \text{ mol}}{10 \text{ L}} = 0.0933 \text{ M} [N2O]0=0 M[N_2O]_0 = 0 \text{ M}

  2. Step 2: Set up the ICE table and equilibrium expression Let 2x2x be the concentration of N2ON_2O formed at equilibrium. At equilibrium: [N2]=0.0482−2x[N_2] = 0.0482 - 2x, [O2]=0.0933−x[O_2] = 0.0933 - x, [N2O]=2x[N_2O] = 2x Since KcK_c is very small, xx is negligible, so [N2]≈0.0482[N_2] \approx 0.0482 and [O2]≈0.0933[O_2] \approx 0.0933.

    Kc=[N2O]2[N2]2[O2]K_c = \frac{[N_2O]^2}{[N_2]^2[O_2]}

  3. Step 3: Substitute values and solve for xx

    2.0×10−37=(2x)2(0.0482)2(0.0933)2.0 \times 10^{-37} = \frac{(2x)^2}{(0.0482)^2(0.0933)} 2.0×10−37=4x20.00021672.0 \times 10^{-37} = \frac{4x^2}{0.0002167} 4x2=4.334×10−414x^2 = 4.334 \times 10^{-41} x2=1.0835×10−41x^2 = 1.0835 \times 10^{-41} x=3.29×10−21 Mx = 3.29 \times 10^{-21} \text{ M}

  4. Step 4: Determine equilibrium concentrations

    [N2O]=2x=2×3.29×10−21=6.58×10−21 M[N_2O] = 2x = 2 \times 3.29 \times 10^{-21} = 6.58 \times 10^{-21} \text{ M} [N2]=0.0482−2(3.29×10−21)≈0.0482 M[N_2] = 0.0482 - 2(3.29 \times 10^{-21}) \approx 0.0482 \text{ M} [O2]=0.0933−3.29×10−21≈0.0933 M[O_2] = 0.0933 - 3.29 \times 10^{-21} \approx 0.0933 \text{ M}
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